Solve 5-|-15|div|3|/2(4-|5|)+9

$$\frac { 5 - | - 15 | \div | 3 | } { 2 ( 4 - | 5 | ) + 9 }$$

Evaluate

$$0$$

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -15 is 15.

$$\frac{5-\frac{15}{|3|}}{2\left(4-|5|\right)+9}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 3 is 3.

$$\frac{5-\frac{15}{3}}{2\left(4-|5|\right)+9}$$

Divide 15 by 3 to get 5.

$$\frac{5-5}{2\left(4-|5|\right)+9}$$

Subtract 5 from 5 to get 0.

$$\frac{0}{2\left(4-|5|\right)+9}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 5 is 5.

$$\frac{0}{2\left(4-5\right)+9}$$

Subtract 5 from 4 to get -1.

$$\frac{0}{2\left(-1\right)+9}$$

Multiply 2 and -1 to get -2.

$$\frac{0}{-2+9}$$

Add -2 and 9 to get 7.

$$\frac{0}{7}$$

Zero divided by any non-zero number gives zero.

$$0$$

Factor

$$0$$